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3. Let
R
be the rate of photon emission (number of photons emitted per unit time) of the
Sun and let
E
be the energy of a single photon. Then the power output of the Sun is given
by
P = RE
. Now
E = hf = hc
/
λ
, where
h
is the Planck constant,
f
is the frequency of the
light emitted, and
λ
is the wavelength. Thus
P = Rhc
/
λ
and
R
P
hc
==
×
×⋅
×
=×
−
λ
550
39 10
6 63 10
2 998 10
10 10
26
34
8
45
nm
W
Js
m
/s
photons/ s.
b
gc
h
c
hc
h
.
..
.
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View Full Document 4. We denote the diameter of the laser beam as
d
. The crosssectional area of the beam is
A =
π
d
2
/4. From the formula obtained in problem 3, the rate is given by
()
(
)
3
2
2
34
8
3
21
2
4 633nm 5.0 10 W
/4
6.63 10
J s
2.998 10 m/s 3.5 10 m
photons
1.7 10
.
ms
RP
A
hc
d
−
−−
×
λ
==
π
π×
⋅
×
×
=×
⋅
18. To find the longest possible wavelength
λ
max
(corresponding to the lowest possible
energy) of a photon which can produce a photoelectric effect in platinum, we set
K
max
= 0
in Eq. 385 and use
hf = hc
/
λ
. Thus
hc
/
λ
max
=
Φ
. We solve for
λ
max
:
λ
max
.
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This note was uploaded on 07/18/2008 for the course PHYS 181 taught by Professor Stephenirons during the Spring '08 term at Yale.
 Spring '08
 STEPHENIRONS
 Physics, Energy, Power, Photon

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